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Quite a straightforward piece of maths I can't seem to get my head around here:

The tangent to the parabola at the point $(4a, 4a)$ is given by what equation?

Bearing in mind the parabola is $y^2 = 4ax$ and the general point $(at^2, 2at)$ lies on the parabola.

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  • $\begingroup$ Do you know about implicit differentiation? You need to find the derivative first $\endgroup$ – Sam Weatherhog Nov 4 '15 at 21:37
  • $\begingroup$ @SamWeatherhog forgive me but what do I differentiate? $\endgroup$ – Pontius Pilate VII Nov 4 '15 at 21:39
  • $\begingroup$ @PontiusPilateVII You differentiate both sides of $y^2=4ax$ with respect to $x$. The left-hand side is done via the chain rule to get $2yy'$. $\endgroup$ – Arthur Nov 4 '15 at 21:42
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Do you know how to compute the tangents to $y=\frac{1}{4a}x^2$ by differentiation? If we take the point $(t,\frac{1}{4a}t^2)$, the tangent has equation $$ y-\frac{1}{4a}t^2=\frac{1}{2a}t(x-t) $$ because the derivative of $x\mapsto \frac{1}{4a}x^2$ is $x\mapsto \frac{1}{2a}x$. For $t=4a$, we get $$ y-\frac{1}{4a}16a^2=\frac{1}{2a}4a(x-4a) $$ that is $$ y-4a=2x-8a $$ or $$ y=2x-4a $$

Just swap $x$ with $y$ and the requested tangent will be $$ x=2y-4a $$

Note that swapping $x$ with $y$ is an isometry, so it preserves tangents.

Let's make another example. Suppose you want to compute the normal to the parabola at the point $(a,-2a)$. You can do it in a very similar way, by computing the normal to the parabola $y=\frac{1}{4a}x^2$ at the point $(-2a,a)$. The tangent will be $$ y-a=\frac{1}{2a}\cdot(-2a)(x+2a) $$ or $$ y-a=-x-2a $$ that is $$ y=-x-a $$ The normal is thus $$ y=x-a $$ so the required normal is $$ x=y-a $$ (by swapping back $x$ with $y$).

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  • $\begingroup$ @PontiusPilateVII If you swap $x$ with $y$ in your given equation, you get $x^2=4ay$, that is, $y=\dfrac{1}{4a}x^2$. $\endgroup$ – egreg Nov 4 '15 at 21:54
  • $\begingroup$ @PontiusPilateVII Where's the difference? An isometry also preserves angles. Don't change the question so that it invalidates existing answers. $\endgroup$ – egreg Nov 4 '15 at 22:26
  • $\begingroup$ Compute the tangent in the “swapped system”; the normal is easy to write; swap back. $\endgroup$ – egreg Nov 4 '15 at 22:29
  • $\begingroup$ @PontiusPilateVII At the end, of course. $\endgroup$ – egreg Nov 4 '15 at 22:32
  • $\begingroup$ @PontiusPilateVII I added the computation. $\endgroup$ – egreg Nov 4 '15 at 22:36

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